TITLE: Automorphisms of rationals
QUESTION [5 upvotes]: Given to me was the following assignment:

Prove that for each $a\in\mathbb Q^{*}$, the mapping from $\mathbb Q$ to $\mathbb Q$ which sends $x$ to $ax$ is an automorphism of the additive group $(\mathbb{Q},+)$. Vice versa, every automotphism of $(\mathbb{Q},+)$ is of this form. Conclude that $\mathrm{Aut}(\mathbb{Q},+)$ is isomorphic to $\mathbb Q^{*}$.

My first problem was that I didn't really 'understand' this problem (still don't). I do understand the definitions. And my idea was that I was asked to basically prove that each homomorphism from $(\mathbb{Q},+)$ to itself can be assigned to a unique $a\in\mathbb Q^{*}$?

REPLY [4 votes]: Let $\phi$ be an automorphism of the additive group  $(\mathbb Q,+)$.
Then $\phi(n) = \phi(1+1+\cdots+1)=\phi(1)+\phi(1)+\cdots+\phi(1)=n\phi(1)$.
Thus, $n\phi(m/n)=\phi(m)=m\phi(1)$ and so $\phi(m/n)= (m/n)\phi(1)$ and $\phi$ is determined by $\phi(1)$.
Since $\phi$ is injective and $\phi(0)=0$, we cannot have $\phi(1)=0$.